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Published byDana Spencer Modified over 6 years ago
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Rational Root Theorem and Fundamental Theorem of Algebra
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Yesterday we were factoring polynomials using synthetic division but we were given a place to start. How would we get started if we were not given a factor?
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You have learned several important properties about real roots of polynomial equations.
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Not all polynomials are factorable, but the Rational Root Theorem can help you find all possible rational roots of a polynomial equation.
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Example 1: Finding All Roots of a Polynomial
Solve x4 – 3x3 + 5x2 – 27x – 36 = 0 by finding all roots. The polynomial is of degree 4, so there are exactly four roots for the equation. Step 1 Use the rational Root Theorem to identify rational roots. ±1, ±2, ±3, ±4, ±6, ±9, ±12, ±18, ±36 p = –36, and q = 1.
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Example 1 Continued Step 2 Graph y = x4 – 3x3 + 5x2 – 27x – 36 to find the real roots. Find the real roots at or near –1 and 4.
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Example 1 Continued Step 3 Test the possible real roots. Test –1. The remainder is 0, so (x + 1) is a factor. –1 1 – –27 –36 –1 4 –9 36 1 –4 9 –36
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Example 1 Continued The polynomial factors into (x + 1)(x3 – 4x2 + 9x – 36) = 0. 4 1 – –36 Test 4 in the cubic polynomial. The remainder is 0, so (x – 4) is a factor. 4 36 1 9
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Example 1 Continued The polynomial factors into (x + 1)(x – 4)(x2 + 9) = 0. Step 4 Solve x2 + 9 = 0 to find the remaining roots. x2 + 9 = 0 x2 = –9 x = ±3i The fully factored form of the equation is (x + 1)(x – 4)(x + 3i)(x – 3i) = 0. The solutions are 4, –1, 3i, –3i.
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You Try
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You try 4. Solve by finding all roots x4 – 5x3 + 7x2 – 5x + 6 = 0
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